Math 484.1 September 29, 2011 Name:_________Dr.V.________________________________
Midterm 1, 5 problems, 15 points each. Return this page with your name on both sides.
1. Solve for x and y where a is a given number:
a2x + y= a2,
ax +ay = 1.
Solution. A row addition operation gives
a2x + y= a2,
(a - a3 )x = 1 - a3.
If a ≠ 0 , ±1, then x = (1 - a3 )/(a - a3 )= (1+ a +a2 )/(a + a2 ) and
y = (1-x)a2 = -a/(1+a).
If a = 0 or -1, then there are no solutions.
If a = 1, then y = 1 - x (x arbitrary).
2. 4x + y 2 -> min,
x2 + y 2 = 10; x and y integers.
Solution There are 8 feasible solutions: (x, y) = (±1, ±3), (±3, ±1).
min =- -11 at x = -3, y = ±1 (two optimal solutions).
3, 4. Solve the linear programs given by the following tableaux with all decision variables xi ≥ 0:
|
x1 |
x2 |
x3 |
-1 |
Problem 3 |
|
-1 |
0 |
-1 |
2 |
= x4 |
|
1 |
0 |
-1 |
-1 |
-> min |
Solution. The standard tableau is
|
x1 |
x2 |
x3 |
1 |
Problem 3 |
|
-1 |
0 |
-1 |
-2 |
= x4 |
|
1 |
0 |
-1 |
1 |
-> min |
The x4-row is bad, so LP is feasible.
|
x1 |
x2 |
-x3 |
1 |
Problem 4 |
|
1 |
0 |
1 |
-2 |
=- x4 |
|
1 |
0 |
-1 |
-1 |
-> min |
Solution. . The standard tableau is
|
x1 |
x2 |
x3 |
1 |
Problem 4 |
|
-1 |
0 |
1 |
2 |
= x4 |
|
1 |
0 |
1 |
-1 |
-> min |
It is optimal so min = -1 at x1 =x2 = x3 = 0, x4 = 2.
5. Find all logical implications between the following 5 constraints on x, y:
(a) x2 = y2, (b) 0 < -2, (c) 0 < 1, (d) x = -y, (e) x=y=0.
Solution.
(b) ⇒ (e) ⇒ (d) ⇒ (a) ⇒ (c) .